# Thread: Integrating absolute values (definite integrals) - can't find good info

1. ## Integrating absolute values (definite integrals) - can't find good info

Anyone have a good tutorial? Google just sends me to people asking questions like I am. I know it's supposed to be split up. Anyways, here is the problem I am stuck on:

from -2 to 2 integrate (1 - 3|x|)dx

Whenever I think about splitting it up on a definite integral, my brain sort of melts. My book is no help either

2. Yes, when dealing with absolute values, break them up.

All it is is two lines. One for negative and one for positive.

$1-3|x|$ is 1+3x and 1-3x

$\int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx$

3. Originally Posted by galactus
Yes, when dealing with absolute values, break them up.

All it is is two lines. One for negative and one for positive.

$1-3|x|$ is 1+3x and 1-3x

$\int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx$
clear, concise

thank you

4. Hello, TYTY!

With absolute values, a sketch usually helps . . .

$\int^2_{-2}\left(1-3|x|\right)\,dx$

We know the graph of $y = |x|$ . . . a V with vertex at the origin.

The graph of $y = 3|x|$ . . . "climbs steeper".

The graph of $y \:=\:-3|x|$ . . . inverts the previous graph.

The graph of $y \:=\:1-3|x|$ . . . moved up one unit.
Code:
              |
*
/|\
/:|:\
-2    /::|::\    2
- - + - o - + - o - + - -
|::/    |    \::|
|:/     |     \:|
|/      |      \|
*       |       *
|

We have an inverted V, vertex at (0,1), x-intercepts at $\pm\tfrac{1}{3}$

They ask for the "area" from $x = \text{-}2$ to $x=2,$
. . and some of it is "negative area".

We don't need Calculus to find the areas of triangles.

[My answer is: -8]

5. Originally Posted by Soroban
Hello, TYTY!

With absolute values, a sketch usually helps . . .

We know the graph of $y = |x|$ . . . a V with vertex at the origin.

The graph of $y = 3|x|$ . . . "climbs steeper".

The graph of $y \:=\:-3|x|$ . . . inverts the previous graph.

The graph of $y \:=\:1-3|x|$ . . . moved up one unit.
Code:
              |
*
/|\
/:|:\
-2    /::|::\    2
- - + - o - + - o - + - -
|::/    |    \::|
|:/     |     \:|
|/      |      \|
*       |       *
|

We have an inverted V, vertex at (0,1), x-intercepts at $\pm\tfrac{1}{3}$

They ask for the "area" from $x = \text{-}2$ to $x=2,$
. . and some of it is "negative area".

We don't need Calculus to find the areas of triangles.

[My answer is: -8]

This is a good point and also how the text wanted it to be done. When I thought about integrating it is where I tripped up though.